If we assume the inner product is linear in the second argument, the polarization identity reads [tex] (x,y) = \frac 14 \| x + y \|^2 - \frac 14 \| x - y \|^2 - \frac i4 \|x + iy\|^2 + \frac i4 \| x - iy \|^2. [/tex] But there is another identity that I've seen referred to in some texts as the "polarization identity", and it is as follows: If T is a linear operator on the Hilbert space, we have [tex] (x,Ty) = \frac 14 (x+y,T(x+y)) - \frac 14 (x-y,T(x-y)) - \frac i4 (x+iy,T(x+iy)) + \frac i4 (x-iy,T(x-iy)). [/tex] I've expanded this out and checked to make sure it's correct, and it is...but is there a quick and painless way to prove this by starting from the first polarization identity above? If so, I really don't see it...my professor seemed to indicate to us that if we defined a kind of ersatz "inner product" on H by letting [itex](x,y)_1 = (x,Ty)[/itex], then the result would follow if we wrote out the first polarization identity with the norm [itex]\| \cdot \|_1[/itex] instead. But is this really kosher? Also, just out of curiosity...why is it appropriate to call this thing a "polarization" identity?